Problem: Let $f(x)=x^3+6x^2+6x$ and let $c$ be the number that satisfies the Mean Value Theorem for $f$ on the interval $[-6,0]$. What is $c$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-5$ (Choice B) B $-4$ (Choice C) C $-3$ (Choice D) D $-1$
According to the Mean Value Theorem, there exists a number $c$ in the open interval $(-6,0)$ such that $f'(c)$ is equal to the average rate of change of $f$ over the interval: $f'(c)=\dfrac{f(0)-f(-6)}{(0)-(-6)}$ First, let's find that average rate of change: $\dfrac{f(0)-f(-6)}{(0)-(-6)}=\dfrac{0-(-36)}{6}={6}$ Now, let's differentiate $f$ and find the $x$ -value for which $f'(x)={6}$. $f'(x)=3x^2+12x+6$ The solutions of $f'(x)=6$ are $x=-4$ and $x=0$. Out of these, only $x=-4$ is within the interval $(-6,0)$. In conclusion, $c=-4$.